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Math Help - i forget a very simple integration problem

  1. #1
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    i forget a very simple integration problem

    i forget a very simple integration problem: is there any way to do this:
    Attached Thumbnails Attached Thumbnails i forget a very simple integration problem-secant-theta.gif  
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  2. #2
    Math Engineering Student
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    \int {\sec \theta \,d\theta } = \int {\frac{{\cos \theta }}<br />
{{(1 + \sin \theta )(1 - \sin \theta )}}\,d\theta } = \frac{1}<br />
{2}\left( {\int {\frac{{\cos \theta }}<br />
{{1 + \sin \theta }}\,d\theta } + \int {\frac{{\cos \theta }}<br />
{{1 - \sin \theta }}\,d\theta } } \right).

    Or just multiply top & bottom by \sec\theta+\tan\theta.
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  3. #3
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    Hello, afeasfaerw!

    \int \sec\theta\,d\theta

    You should have been taught the formula for this by now . . .

    It is usually derived like this . . .


    Multiply by . \frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}

    . . \int\frac{\sec\theta}{1}\cdot\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec^2\!\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec\theta\tan\theta + \sec^2\!\theta}{\sec\theta + \tan\theta}\,d\theta


    Let: u \:=\:\sec\theta + \tan\theta\quad\Rightarrow\quad du \:=\:(\sec\theta\tan\theta + \sec^2\theta)d\theta

    Substitute: . \int\frac{du}{u}\;=\;\ln|u| + C

    Back-substitute: . \ln|\sec\theta + \tan\theta| + C

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